[LeetCode]14. Longest Common Prefix最长公共前缀

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Example 1:

Input: ["flower","flow","flight"]
Output: "fl"

Example 2:

Input: ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.

Note:

All given inputs are in lowercase letters a-z.

求几个字符串的公共前缀，两种方法

(1)暴力

利用两层循环，先取出第一个字符串的第i个字符，然后比较s[1]-s[n]的第i个字符是否相等

public class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs == null || strs.length == 0) return "";
        String res ="";
        for (int i = 0; i < strs[0].length(); i++) {
            char c = strs[0].charAt(i);
            for (int j = 1; j < strs.length; j++) {
                if (i >= strs[j].length() || strs[j].charAt(i) != c) {
                    return res;
                }
            }
            res =res+c;
        }
        return res;
    }
}

(2)排序后比较首尾

字符串比较的结果，按照首个不同字符的大小排序，比如ab就排在ac的前面

["flower","flow","flight"]比较的结果是[flight,flow,flower]，我们统计首尾，fl是最长前缀

["dog","racecar","car"]比较的结果是[car,dog,racecar]
根据这个特性，假如存在最长公共前缀，s[0]和s[n]之间的最长公共前缀就是我们要求的结果，因为排序结果是按照第一个非公共字符开始的
这里我们把过程简化到只要比较首尾字符，而不用比较所有的

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs == null || strs.length == 0) return "";
        Arrays.sort(strs);
        int i = 0, len = Math.min(strs[0].length(), strs[strs.length - 1].length());
        while (i < len && strs[0].charAt(i) == strs[strs.length - 1].charAt(i)) i++;
        return strs[0].substring(0, i);
    }
}